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A baseball player (Fig. 5-31) with mass \(79 \mathrm{~kg}\), sliding into a base, is slowed by a force of friction of \(470 \mathrm{~N}\). What is the coefficient of kinetic friction between the player and the ground?

Short Answer

Expert verified
The coefficient of kinetic friction between the player and the ground is approximately 0.61.

Step by step solution

01

Calculate the normal force (Fn)

The normal force (Fn) is the weight of the baseball player which can be calculated by multiplying mass with gravity. Here, the mass is \(79 \mathrm{~kg}\) and gravity is \(9.8 \mathrm{~m/s^2}\). So, \(Fn = mass \times gravity = 79 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2} = 774.2 \mathrm{~N}\)
02

Calculate the coefficient of kinetic friction

Now the coefficient of kinetic friction can be calculated with the formula \(\mu = F_f / F_n \), where the frictional force \(F_f\) is given as \(470 \mathrm{~N}\). So, \(\mu = 470 \mathrm{~N} / 774.2 \mathrm{~N} = 0.6072\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Force
The normal force is a vital concept in understanding how objects interact with surfaces. It is essentially the support force exerted by a surface, perpendicular to the surface. When an object is placed on a flat surface, the normal force balances the object's weight due to gravity, keeping it stationary.In our baseball player scenario, the normal force acts upwards, counteracting the player's weight. To compute this, we multiply the player's mass by the acceleration due to gravity. Given:
  • Mass of player: 79 kg
  • Gravity: 9.8 m/s²
The normal force (Fn) is calculated as:\[ F_n = \text{mass} \times \text{gravity} = 79 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 774.2 \, \text{N} \]This tells us that the force pushing up against the player, due to the ground, is 774.2 N. Understanding normal force is crucial as it is a key player in calculating friction.
Force of Friction
Friction is the force resisting the relative motion of surfaces sliding past each other. It is critical to many everyday activities such as walking or driving. There are two main types of friction: static and kinetic. Our focus here is kinetic friction because our baseball player is sliding. The force of friction ( F_f ) acts opposite to the direction of motion. In this exercise, friction slows the player down as they slide into the base. It is given as 470 N. Kinetic friction is dependent on:
  • The normal force
  • The coefficient of kinetic friction
Together, they determine how much the surface impedes motion. Remember, while friction can slow things down, it is also essential for controlling movement which is incredibly helpful in various sports and applications.
Coefficient of Friction
The coefficient of kinetic friction (\mu_k) is a dimensionless number that represents how easily one surface slides over another. It varies depending on the textures and materials involved.Calculating \mu_k is straightforward once the normal force and the frictional force are known. The formula is:\[ \mu_k = \frac{F_f}{F_n} \]Substituting the values from our baseball scenario:
  • Frictional force (F_f): 470 N
  • Normal force (F_n): 774.2 N
We find:\[ \mu_k = \frac{470 \, \text{N}}{774.2 \, \text{N}} \approx 0.6072 \]This tells us that the surfaces (player and ground) aren’t perfectly smooth, indicating a moderate level of friction. By understanding \mu_k, you can predict how different surfaces will interact under similar conditions.

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Most popular questions from this chapter

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}(=37 \mathrm{mi} / \mathrm{h}) .(a)\) If the radius of the curve is \(150 \mathrm{~m}\) \((=490 \mathrm{ft})\), what is the correct angle of banking of the road? ( \(b\) ) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding at this speed?

An elevator weighing \(6200 \mathrm{lb}\) is pulled upward by a cable with an acceleration of \(3.8 \mathrm{ft} / \mathrm{s}^{2}\). (a) What is the tension in the cable? (b) What is the tension when the elevator is accelerating downward at \(3.8 \mathrm{ft} / \mathrm{s}^{2}\) but is still moving upward?

The position of a particle of mass \(2.17\) kg traveling in a straight line is given by $$ x=\left(0.179 \mathrm{~m} / \mathrm{s}^{4}\right) t^{4}-\left(2.08 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}+17.1 \mathrm{~m} $$ Find the \((a)\) velocity, \((b)\) acceleration, and \((c)\) force on the particle at time \(t=7.18 \mathrm{~s}\)

Assume that the standard kilogram would weigh exactly \(9.80 \mathrm{~N}\) at sea level on the equator if the Earth did not rotate. Then take into account the fact that the Earth does rotate, so that this object moves in a circle of radius \(6370 \mathrm{~km}\) (the Earth's radius) in one day. (a) Determine the centripetal force needed to keep the standard kilogram moving in its circular path. ( \(b\) ) Find the force exerted by the standard kilogram on a spring balance from which it is suspended at the equator (its apparent weight).

An \(11-\mathrm{kg}\) monkey is climbing a massless rope attached to a \(15-\mathrm{kg}\) log over a frictionless tree limb. (a) With what minimum acceleration must the monkey climb up the rope so that it can raise the \(15-\mathrm{kg}\) log off the ground? If, after the \(\log\) has been raised off the ground, the monkey stops climbing and hangs on to the rope, what will now be \((b)\) the monkey's acceleration and \((c)\) the tension in the rope?

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